Question

1. find the measures of the angles in the figure. 50°, 50°, 100°, 100° 60°, 60°, 120°, 120° 50°, 50°, 130°, 130° 65°, 65°, 115°, 115°

Explanation:

Step1: Recall the sum of angles in a quadrilateral

The sum of the interior angles of a quadrilateral is \(360^\circ\).

Step2: Set up the equation for the trapezoid

The trapezoid has two angles of \(h^\circ\) and two angles of \(2h^\circ\). So the sum of the angles is \(h + h + 2h + 2h = 360\).

Step3: Simplify and solve for \(h\)

Combining like terms, we get \(6h = 360\). Dividing both sides by 6, we have \(h=\frac{360}{6}=60\). Wait, no, wait, let's check again. Wait, no, the trapezoid here: wait, actually, a trapezoid (isosceles trapezoid) has consecutive angles supplementary? Wait, no, the sum of interior angles of a quadrilateral is \(360^\circ\). So the angles are \(h, h, 2h, 2h\). So total sum: \(h + h + 2h + 2h = 6h = 360\), so \(h = 60\)? But wait, let's check the options. Wait, no, maybe I made a mistake. Wait, no, let's recalculate. Wait, \(h + h + 2h + 2h = 6h\), so \(6h = 360\), so \(h = 60\), then the angles would be \(60^\circ, 60^\circ, 120^\circ, 120^\circ\), which is the second option. Wait, but let's check again. Wait, maybe the trapezoid is not isosceles? No, the figure shows two angles of \(h\) and two of \(2h\), so it's isosceles. So sum is \(6h = 360\), so \(h = 60\), so angles are \(60, 60, 120, 120\). Wait, but let's check the options. The second option is \(60^\circ, 60^\circ, 120^\circ, 120^\circ\). Wait, but wait, maybe I messed up the sum. Wait, no, quadrilateral sum is \(360\). So \(h + h + 2h + 2h = 6h = 360\), so \(h = 60\). So the angles are \(h = 60\), \(2h = 120\). So two angles of \(60^\circ\) and two of \(120^\circ\), which is the second option. Wait, but let's check the other options. Wait, maybe I made a mistake. Wait, no, let's check again. Wait, the first option: \(50 + 50 + 100 + 100 = 300\), no. Second: \(60 + 60 + 120 + 120 = 360\), yes. Third: \(50 + 50 + 130 + 130 = 360\)? Wait, \(50+50=100\), \(130+130=260\), total \(360\). Wait, that also sums to \(360\). Wait, so my initial equation was wrong? Wait, no, the figure has angles \(h, h, 2h, 2h\), so the ratio is \(1:1:2:2\). So the angles should be in the ratio \(1:1:2:2\). Let's check the ratios of the options. First option: \(50:50:100:100 = 1:1:2:2\). Second: \(60:60:120:120 = 1:1:2:2\). Third: \(50:50:130:130 = 5:5:13:13\), not \(1:1:2:2\). Fourth: \(65:65:115:115 = 13:13:23:23\), not \(1:1:2:2\). Wait, so both first and second options have ratio \(1:1:2:2\). Wait, so my mistake was in the sum? Wait, no, \(50 + 50 + 100 + 100 = 300\), which is not \(360\). Wait, \(50*2 + 100*2 = 100 + 200 = 300\), which is wrong. \(60*2 + 120*2 = 120 + 240 = 360\), which is correct. So the first option's sum is \(300\), which is wrong. So the second option is correct. So my initial calculation was right. So \(h = 60\), \(2h = 120\). So the angles are \(60^\circ, 60^\circ, 120^\circ, 120^\circ\).

Answer:

B. \(60^\circ, 60^\circ, 120^\circ, 120^\circ\) (assuming the options are labeled A, B, C, D with B being the second option as per the order: first option is 50,50,100,100; second is 60,60,120,120; etc.)